# Proving $\frac{1}{n} \sum_{k=1}^{n}X_{k}\to 0$ a.s.

almost-everywhereborel-cantelli-lemmasprobability theory

$$\{X_{n}\}$$ is a sequence of independent random variables, $$EX_n=0$$, and $$\sum_{n=1}^{\infty}n^{-(r+1)}E(|X_n|^{2r})<\infty$$. Proving $$\frac{1}{n} \sum_{k=1}^{n}X_{k}\to 0$$ a.s. and $$r>1$$

I think Borel-Cantelli lemma should be a very useful way to prove this kind of problem, but I don’t know how it should be applied to this one. Some approaches are welcome!

It suffices to prove (in view of the Borel-Cantelli lemma and the fact that $$a_n/n\to 0$$ if and only if $$\max_{1\leqslant n\leqslant 2^N}a_n/2^N\to 0$$) that for each positive $$\varepsilon$$, $$\sum_{N\geqslant 1}\mathbb P\left(\max_{1\leqslant n\leqslant 2^N}\left\lvert \sum_{k=1}^nX_k\right\rvert >\varepsilon 2^N\right)<\infty.$$ By Markov's inequality, we are reduced to establish the convergence of $$\sum_{N\geqslant 1}b_N$$, where $$b_N:=2^{-2Nr}\mathbb E\left[\max_{1\leqslant n\leqslant 2^N}\left\lvert \sum_{k=1}^nX_k\right\rvert^{2r}\right].$$ Since $$2r>1$$, Doob's inequality gives $$b_N\leqslant \left(\frac{2r}{2r-1}\right)^{2r-1}2^{-2Nr}\mathbb E\left[ \left\lvert \sum_{k=1}^{2^N}X_k\right\rvert^{2r}\right].$$ Using Marcinkiewicz–Zygmund inequality with $$p=2r$$, we get $$b_N\leqslant C_r2^{-Nr}\mathbb E\left[ \left(2^{-N} \sum_{k=1}^{2^N}X_k^2\right)^{r}\right],$$ where $$C_r$$ depends only on $$r$$. Since $$r>1$$, we get by convexity of $$t\mapsto t^r$$ that $$b_N\leqslant C_r2^{-Nr} 2^{-N}\sum_{k=1}^{2^N}\mathbb E\left[\lvert X_k\rvert^{2r}\right] .$$ As a consequence, $$\sum_N b_N$$ converges as long as $$\sum_N2^{-Nr} 2^{-N}\sum_{k=1}^{2^N}\mathbb E\left[\lvert X_k\rvert^{2r}\right]$$ does, that is, $$\sum_{k\geqslant 1}\sum_{N: 2^N\geqslant k}2^{-N(r+1)}\mathbb E\left[\lvert X_k\rvert^{2r}\right]<\infty,$$ which is exactly the mentioned condition because $$\sum_{N: 2^N\geqslant k}2^{-N(r+1)}$$ is of order $$k^{-r-1}$$.
As pointer out in the comments, it is not required that the $$X_k$$ share the same distribution.